From 47d7279a724246ef7aa0f5359cf417992ed04449 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Wed, 3 Jun 2015 15:27:17 +0530 Subject: add books --- sample_notebooks/ChandraShiva/Chapter_1.ipynb | 241 ++++++++++++++++++++++++++ 1 file changed, 241 insertions(+) create mode 100755 sample_notebooks/ChandraShiva/Chapter_1.ipynb (limited to 'sample_notebooks/ChandraShiva') diff --git a/sample_notebooks/ChandraShiva/Chapter_1.ipynb b/sample_notebooks/ChandraShiva/Chapter_1.ipynb new file mode 100755 index 00000000..7bea3912 --- /dev/null +++ b/sample_notebooks/ChandraShiva/Chapter_1.ipynb @@ -0,0 +1,241 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:5f4c2f0ce05563ebaa82d1579fc74f4b0a66c01a8b91c63f65c05e0494b44f20" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1 - Introduction to Electronics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E1 - Pg 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Find the range of tolerance\n", + "#soltion\n", + "#given\n", + "#color coding\n", + "orange=3.#\n", + "gold=5.#\n", + "yellow=4.#\n", + "violet=7.#\n", + "#band pattern\n", + "band1=yellow#\n", + "band2=violet#\n", + "band3=orange#\n", + "band4=gold#\n", + "#resistor color coding\n", + "r=(band1*10.+band2)*10.**(band3)#\n", + "tol=r*(band4/100.)#tolerance\n", + "ulr=r+tol##upper limit of resistance\n", + "llr=r-tol##lower limit of resistance\n", + "print 'The range of resistance =',llr/1000. ,'kOhm','to',ulr/1000,'kOhm'\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The range of resistance = 44.65 kOhm to 49.35 kOhm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E2 - Pg 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Find the range of tolerance\n", + "#color coding\n", + "blue=6.#\n", + "gold=-1.#\n", + "gray=8.#\n", + "silver=10.#\n", + "#band pattern\n", + "band1=gray#\n", + "band2=blue#\n", + "band3=gold#\n", + "band4=silver#\n", + "#resistor color coding\n", + "r=(band1*10.+band2)*10.**(band3)#\n", + "tol=r*(band4/100.)#tolerance\n", + "ulr=r+tol##upper limit of resistance\n", + "llr=r-tol##lower limit of resistance\n", + "print 'The Range of resistance is',llr,'ohm','to',ulr,'ohm'\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Range of resistance is 7.74 ohm to 9.46 ohm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E3 - Pg 19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Find the equivalent current source\n", + "#given\n", + "Vs=2.;#Volts #dc voltage source\n", + "Rs=1.;#ohm #internal resistance\n", + "Rl=1.;#ohm #load resistance\n", + "Ise=Vs/Rs;#ampere #equivalent current source\n", + "\n", + "# In accordance to figure 1.23a\n", + "Il1=Ise*(Rs/(Rs+Rl));#using current divider concept\n", + "Vl1=Il1*Rl;\n", + "print \"In accordance to figure 1.23a\\n\"\n", + "print \"The Load current (current source Il=\",Il1,'A'\n", + "print \"The Load voltage (current source Vl=\",Vl1,'V','\\n'\n", + "\n", + "# In accordance to figure 1.23b\n", + "Vl2=Vs*(Rs/(Rs+Rl));#using voltage divider concept\n", + "Il2=Vl2/Rl;\n", + "print \"\\nIn accordance to figure 1.23b\"\n", + "print \"\\nThe Load voltage (voltage source) Vl=\",Vl2,'V'\n", + "print \"The Load current (voltage source) Il=\",Il2,'A'\n", + "print \"\\nTherefore they both provide same voltage and current to load\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "In accordance to figure 1.23a\n", + "\n", + "The Load current (current source Il= 1.0 A\n", + "The Load voltage (current source Vl= 1.0 V \n", + "\n", + "\n", + "In accordance to figure 1.23b\n", + "\n", + "The Load voltage (voltage source) Vl= 1.0 V\n", + "The Load current (voltage source) Il= 1.0 A\n", + "\n", + "Therefore they both provide same voltage and current to load\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E4 - Pg 19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Find percentage variation in load current and load voltage\n", + "#given\n", + "Vs=10.;#volt#Supply voltage\n", + "Rs=100.;#ohm#internal resistance\n", + "\n", + "# In accordance to figure 1.24a\n", + "#For 1ohm - 10 ohm\n", + "Rl11=1.;#ohm#min extreme value of Rl\n", + "Rl12=10.;#ohm#max extreme value of Rl\n", + "Il11=Vs/(Rs+Rl11);\n", + "Il12=Vs/(Rs+Rl12);\n", + "Pi1=(Il11-Il12)*100./Il11;#Percentage variation in current\n", + "Vl11=Il11*Rl11;\n", + "Vl12=Il12*Rl12;\n", + "Pv1=(Vl12-Vl11)*100./Vl12;#Percentage variation in voltage\n", + "print '%s' %(\"In accordance to figure 1.24a \\n\");\n", + "print '%s %.2f %s' %(\"Percentage variation in Current(1-10 ohm)=\",Pi1,'percent');\n", + "print '%s %.1f %s ' %(\"Percentage variation in Voltage(1-10 ohm)=\",Pv1,'percent\\n\\n');\n", + "\n", + "# In accordance to figure 1.24b\n", + "#For 1kohm - 10kohm\n", + "Rl21=1000.;#ohm#min extreme value of Rl\n", + "Rl22=10000.;#ohm#max extreme value of Rl\n", + "Il21=Vs/(Rs+Rl21);\n", + "Il22=Vs/(Rs+Rl22);\n", + "Pi2=(Il21-Il22)*100./Il21;#Percentage variation in current\n", + "Vl21=Il21*Rl21;\n", + "Vl22=Il22*Rl22;\n", + "Pv2=(Vl22-Vl21)*100./Vl22;#Percentage variation in voltage\n", + "print '%s' %(\"In accordance to figure 1.24b \\n\");\n", + "print '%s %.f %s' %(\"Percentage variation in Current(1-10 ohm)=\",Pi2,'percent');\n", + "print '%s %.f %s ' %(\"Percentage variation in Voltage(1-10 ohm)=\",Pv2,'percent \\n');\n", + "print 'In book the percentage variation in voltage(1kohm-10kohm) is 9 percent due to' \n", + "print 'the incorrect value of Il22 i.e. 0.000999 Amp correct value is 0.0009901 Amp'\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "In accordance to figure 1.24a \n", + "\n", + "Percentage variation in Current(1-10 ohm)= 8.18 percent\n", + "Percentage variation in Voltage(1-10 ohm)= 89.1 percent\n", + "\n", + " \n", + "In accordance to figure 1.24b \n", + "\n", + "Percentage variation in Current(1-10 ohm)= 89 percent\n", + "Percentage variation in Voltage(1-10 ohm)= 8 percent \n", + " \n", + "In book the percentage variation in voltage(1kohm-10kohm) is 9 percent due to\n", + "the incorrect value of Il22 i.e. 0.000999 Amp correct value is 0.0009901 Amp\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit